Why does $a^{\log_a(x)}=x$? I'm currently learning about logs and yesterday made a post on here.
In that post I was told that $a^{\log_a(x)}=x$
From a commenter:

"It should be clear that..."
Nope, not for me. I've tried coming back to this since posting yesterday but cannot 'see it' or make it click. I'm seeking hand holding and a low level response. Why does $a^{log_a(x)}=x$?
 A: By definition, the $\log-$function is the inverse of the exponential function. It means that, if $f:\mathbb R\to \mathbb R^+$ such that,
$$f(x)=a^x$$
then its inverse is a function $f^{-1}:\mathbb R^+\to \mathbb R$ such that,
$$f(f^{-1}(x))=x.$$
We then define $f^{-1}$ as $f^{-1}(x)=\log_a x$. So,
$$f(f^{-1}(x))=x\Leftrightarrow a^{\log_a x}=x.$$
A: It is true that by definition $a^{\log_a(x)}=x$. However, perhaps it would be worthwhile to show you the motivation behind why we define logarithms in this way. As @Stinking Bishop has already pointed out, logarithms help us solve for $x$ when dealing with an equation of the form
$$
a^x=b \, .
$$
For example, if
$$
10^x=100
$$
then $x=2$ is the solution. But suppose we were trying to solve
$$
10^x=101 \, .
$$
Now, it is not immediately clear what $x$ is. We can approximate its decimal value using trial and error, but this is only an approximation. In the same way that $\sqrt{2}$ represents the exact positive solution of $x^2=2$,
$$
\log_{10}(101)
$$
represents the exact solution to the equation
$$
10^x=101 \, .
$$
This means that
$$
10^{\log_{10}(101)}=101
$$
by definition. $\log_{10}(101)$ is the 'label' we give to the solution of
$$
10^x=101 \, .
$$
More generally, if $a^x=b$, then $\log_a{b}$, by virtue of how logarithms are defined, represents the solution of the equation. Thus,
$$
a^{\log_a{b}}=b
$$
by definition. Now, a more sophisticated way of thinking about logarithms is that they inverse exponentiation. If $f(x)=a^x$, then $f^{-1}(x)=\log_a(x)$. The defining feature of inverse functions is that $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ for all $x$. Hence,
$$
\log_a(a^x) = x \text{ and } a^{\log_a{x}}=x
$$
are both true by definition. The hardest part is trying to explain why these two conceptions of logarithms, while superficially different, are actually the same. Let's return to our equation
$$
a^x=b \, .
$$
If we 'take logs of both sides', we get
$$
\log_a(a^x) = \log_a(b) \, ,
$$
which simplifies to
$$
x= \log_a(b).
$$
Thus, by defining the logarithm as the inverse of the exponential, $x$ represents the power that we must raise $a$ to in order to get $b$. And this should make intuitive sense. In general, the inverse function should tell us 'how to go backwards'. If we start with the number $x$ and perform a function on it so that we get $f(x)$, then $f^{-1}$ is like an instruction manual for how to get back to $x$. Thus,
$$
f^{-1}(f(x))=x \, .
$$
In the case of exponentiation, we start with a number with a number $x$, and raise $a$ to the power of $x$ to get $a^x$. The logarithm, being the inverse function, should let us return to $x$:
$$
\log_a(a^x)=x \, .
$$
Also, if we first perform $f^{-1}$ on $x$, then again, $f$ is the inverse function. Try working out what this means in the context of exponentials and logarithms.
A: Consider the statement $$\color{red}{2}^\color{green}{3}=\color{blue}{8}$$
You can say that:

*

*$\color{green}{3}=\log_\color{red}{2}(\color{blue}{8})$

*$\color{red}{2}=\sqrt[\color{green}{3}]{\color{blue}{8}}$
Instead of asking "what number must $\color{red}{2}$ be raised to, to give $\color{blue}{8}$" you may ask "what is $\log_\color{red}{2}(\color{blue}{8})$?". You can then consider $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{x})}$$
$\log_\color{red}{a}(\color{blue}{x})$ is "the number that gives $\color{blue}{x}$ when $\color{red}{a}$ is raised to it", so Очевидно, что $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{x})}=\color{blue}{x}$$

In general (given that both $\log_a(c), \sqrt[b]c$ exist) if $$\color{red}{a}^\color{green}{b}=\color{blue}{c}$$ then

*

*$\color{green}{b}=\log_\color{red}{a}(\color{blue}{c})$, which means that $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{c})}=\color{blue}{c}$$


*$\color{red}{a}=\sqrt[\color{green}{b}]{\color{blue}{c}}$, which means that $$(\sqrt[\color{green}{b}]{\color{blue}{c}})^{\color{green}{b}}=\color{blue}{c}$$
I recommend watching this.
A: Assume that $a^b = x$. If we take the logarithm of both sides in base $a$, we will get $\log_a(a^b) = \log_a(x)$. By the definition of logarithm, $log_a(a^b) = b$, since logarithm function returns the exponent of an input number in the given base. Thus, returning to $\log_a(a^b) = \log_a(x)$, we can see that $b = \log_a(a^b) = \log_a(x)$, and $b = \log_a(x)$. Since our original equation is  $a^b = x$, we can just substitute $b$ in the original equation to find the expression $a^{\log_a(x)} = x$. Hope this is clear!
